In finding cyclic subgroups, does it suffice to check one element generators?

In particular, I'm trying to find the cyclic subgroups of $D_8=\{1,r,r^2,r^3,s,sr,sr^2,sr^3\}$, the dihedral group. Seems like quite a hassle to check every possible subset of $D_8$ as a generator. I've gone through each one element generator to see what cyclic subgroups I get from that.

Is there any time-saving insight that says it's good enough to check the one element generators (in general, or just for this problem)?

Thanks!

• What specifically do you mean by 'one-element generators'? It's true that every cyclic group is generated by a single element, so you only have to check the individual elements to see what groups they generate - but in the case you give, for instance, that doesn't mean that you only have to test $r$ and $s$ as generators... – Steven Stadnicki Feb 15 '17 at 23:33
• I meant I checked each element of $D_8$. Ok, I missed that in reading the definition of cyclic groups. Thanks a bunch! – manofbear Feb 15 '17 at 23:34
• One way to save further work is you know the only candidates besides the trivial group are $Z_2$ and $Z_4$ since the order of the group is $8$ and Lagrange's thm implies the order of the subgroups must be $1,2,4,$ or $8.$ $Z_8$ is out cause $D_8$ has $8$ elements and is not $Z_8.$ Then you just find one element that generates $Z_2$ and one that generates $Z_4$ and don't have to bother checking any of the others. – spaceisdarkgreen Feb 15 '17 at 23:39
• @space While you're correct that those are the only possible isomorphism classes of cyclic subgroups (except for the trivial group), there are $5$ copies of $\mathbb Z_2$ and I believe the goal is to find each specific one. In that case you still need to check every element. – Matt Samuel Feb 15 '17 at 23:42
• @MattSamuel true, I think that's probably the correct interpretation of the question. – spaceisdarkgreen Feb 15 '17 at 23:47